Ring 1 : $ \{a+b\sqrt{3}\}: a,b \in \mathbb{Z} $
Ring 2 : $ \{a+b\sqrt{3}\}: a,b \in \mathbb{Q} $
Both clearly commutative with identity.
Need to check that for each of the rings they both have a multiplicative inverse.
Also need to show theres no zero divisors.
I understand both of these but not sure how to show it. Thanks
On
Expand $(a+b\sqrt3)(x+y\sqrt3)=1$ and hence find $x,y$ in terms of $a,b$. Thus the inverse exists if $a\ne0$ or $b\ne0$ (and consists of rational coefficients). It then follows that if $(a+b\sqrt3)(c+d\sqrt3)=0$ then either $a=b=0$ or else $a+b\sqrt3$ is invertible, so $c=d=0$.
On
For $R_1 := \{a+b\sqrt{3} \ | \ a,b \in \mathbb{Z}\}$, the element $2 \in R_1$ has no multiplicative inverse in $R_1$ since $\dfrac{1}{2} \not\in R_1$.
For $R_2 := \{a+b\sqrt{3} \ | \ a,b \in \mathbb{Q}\}$, any element $a+b\sqrt{3} \in R_2$ has multiplicative inverse $\dfrac{a-b\sqrt{3}}{a^2-3b^2} \in R_2$.
Both of them have no zero divisor since they are subrings of $\mathbb{R}.$
Ring $1$ is $\mathbb Z[\sqrt 3]\cong \mathbb Z[X]/(X^2-3)$ and ring 2 is $\mathbb Q(\sqrt 3)\cong \mathbb Q[X]/(X^2+3)$. Then $\mathbb Q(\sqrt 3)$ is clearly a field (because $(X^2+3)$ is maximal), and $\mathbb Z[\sqrt 3]$ is a domain (because $X^2+3$ is prime).